60 lines
1.7 KiB
Markdown
60 lines
1.7 KiB
Markdown
---
|
||
title: leetcode-121
|
||
date: 2020-03-17 18:06:45
|
||
tags:
|
||
categories: leetcode
|
||
---
|
||
|
||
### 121. 买卖股票的最佳时机
|
||
|
||
[题目](https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock/)
|
||
|
||
<!--more-->
|
||
|
||
原始答案：
|
||
|
||
```python
|
||
class Solution:
|
||
def maxProfit(self, prices) -> int:
|
||
profit = 0
|
||
if not prices:
|
||
return profit
|
||
min_buyin = prices[0]
|
||
max_sellout = prices[0]
|
||
l = len(prices)
|
||
i = 0
|
||
for buyin in prices:
|
||
if i == l:
|
||
return profit
|
||
if min_buyin <= buyin:
|
||
max_sellout = max(prices[i:])
|
||
p = max_sellout - min_buyin
|
||
if p > profit:
|
||
profit = p
|
||
if buyin < min_buyin:
|
||
min_buyin = buyin
|
||
i += 1
|
||
return profit
|
||
#1880 ms 14.4 MB
|
||
```
|
||
|
||
主要思路是找到波谷，如果当前价格比前一天要低，则还是在去往波谷的路上；当价格比前一天高或相同时，则到达了一个波谷，计算波谷和之后的波峰的差，就是这一段的利润。将从头至尾过一次，就能找到所有波谷和其后波峰的差，返回最大的即可。但是这个明显地在重复max，时间复杂度是O(n^2)，看起来就很傻逼。
|
||
|
||
仔细想想，其实并不需要直接找出波谷后的波峰，只要在for loop时保持波谷为最低的那个，就能算出每一个后续与波谷的差，找最大差即可。改了下代码变成这样
|
||
|
||
```python
|
||
class Solution:
|
||
def maxProfit(self, prices) -> int:
|
||
profit = 0
|
||
if not prices:
|
||
return profit
|
||
min_buyin = prices[0]
|
||
for i in range(1, len(prices)):
|
||
min_buyin = min(min_buyin, prices[i-1])
|
||
profit = max(prices[i] - min_buyin, profit)
|
||
return profit
|
||
#44 ms 14.4 MB
|
||
```
|
||
|
||
|